AP Chemistry TEST 6 Branham High School Exam 6: Chapters 8&9 Ms. Barkley SECTION 1: Multiple-choice (25 minutes, No Calculators)
Directions: Each set of lettered choices below refers to the numbered statements immediately following it. Select the one lettered choice that best fits each statement and then blacken the
corresponding space on the answer sheet. A choice may be used once, more than once, or not at all in each set.
1. There are ________ shared and ________ unpaired electrons in the Lewis Structure for PH3.
A. 4, 2 B. 2, 4 C. 3, 2 D. 4, 3 E. 2, 3
2. Based on the octet rule, magnesium most likely forms a ______ ion A. Mg2+B. Mg2- C. Mg6- D. Mg6+E. Mg
-3. The electron configuration of the phosphide ion (P3-) is ___________. A. [Ne]3s2
B. [Ne]3s23p1 C. [Ne]3s23p3 D. [Ne]3p2 E. [Ne]3s23p6
4. The chloride of which of the following metals should have the greatest lattice energy? A. potassium B. rubidium C. sodium D. lithium E. cesium
5. The principal quantum number of the electrons that are lost when tungsten forms a cation is ___________.
A. 6 B. 5 C. 4 D. 3 E. 2
6. Which energy change corresponds to the negative lattice energy of potassium fluoride? (see the Born-Haber cycle diagram below)
A. 2 B. 5 C. 4 D. 1 E. 6
K(g) + 1/2F2(g)
K(s) + 1/2F2(g)
K(s) + F(g)
K+
(g) + F(g)
K+
(g) + F-(g)
KF(s)
1 2 3
4
6
7. Which energy change corresponds to the first ionization energy of potassium? (see the diagram of the Born-Haber cycle from the previous problem)
A. 2 B. 5 C. 4 D. 3 E. 6
8. In which of the molecules below is the carbon-carbon distance the shortest? A. H2C=CH2
B. H-C = C-H C. H3C-CH3
D. H2C=C=CH2
E. H3C-CH2-CH3
9. In the molecule below, which atom has the largets partial negative charge? A. Cl
B. F C. Br D. I E. C
10. Given the electronegativities below, which covalent bond is the most polar?
Element: H C N O
Electronegativity: 2.1 2.5 3.0 3.5
A. C-H B. N-H C. O-H D. O-C E. O-N
11. Which of the following names is the correct name for the compound ZnO?
A. zinc (II) dioxide B. zinc dioxide C. zinc oxide D. zinc (IV) oxide E. zinc (II) oxide
12. Which Lewis Structure has the largest TOTAL number of loan pairs of electrons?
A. H2O B. NH3 C. HF D. BHCl2 E. CO2
13. Calculate the formal charge on the carbon atom in carbon dioxide.
A. -2 B. 1 C. 2 D. 0 E. -1
C F
I Cl
14. In the nitrite ion (NO2-), _________.
A. both bonds are single bonds B. both bonds are double bonds
C. one bond is a double bond and the other is a single bond D. both bonds are the same
E. there are 20 valence electrons
15. Resonance structures differ by ___________. A. number and placement of electrons
B. number of electrons only C. placement of atoms only D. placement of electrons only E. number of atoms only
16. Bond enthalpy is __________. A. always zero
B. always negative
C. sometimes positive, sometimes negative D. always positive
E. unpredictable
17. According to VSEPR theory, if there are three places where electrons can be found around a central atom, the geometry will be __________.
A. octahedral B. linear C. bent
D. trigonal planar E. tetrahedral
18. When counting places where electrons can be found around a central atom, a _________ is not included.
19. The molecular geometry of water is __________.
A. linear B. tetrahedral C. octahedral D. bent E. trigonal planar
20. The F-B-F bond angle in the BF2 - ion is approximately ___________.
A. 90o B. 109.5o C. 180o D. 120o E. 60o
21. According to VSEPR Theory, if there are two places where electrons can be found on a central atom, they will be arranged such that the angels between the domains are ____________.
A. 360o B. 120o C. 109.5o D. 180o E. 90o
22. Of the molecules below, only _________ is nonpolar.
A. CO2 B. H2O C. NH3 D. HCl E. TeCl2
23. Of the following, only ___ has sp2 hybridization.
A. PH3 B. CO32- C. ICl D. H2S E. PF5
24. The sp3d atomic hybrid orbital set accommodates _____________ bonds.
A. 4 B. 3 C. 5 D. 6 E. 2
25. The pi bond in ethylene, CH2CH2, results from the overlap of __________.
A. sp3 hybrid orbitals
B. s atomic orbitals C. so hybrid orbitals D. sp2 hybrid orbitals
E. p atomic orbitals
26. Valence bond theory does not address the issue of __________.
A. excited states of molecules B. molecular shapes C. covalent bonding
D. hybridization E. multiple bonds
27. Based on molecular orbital theory, the only molecule in the list below that has unpaired electrons is ____________.
28. Based on molecular orbital theory, the bond order of the N-N bond in the N2 is _______.
A. 0 B. 2 C. 1 D. 3 E. 1/2
29. According to molecular orbital theory, overlap of two atomic orbitals produces _____________. A. one bonding molecular orbital and one hybrid orbital
B. two bonding molecular orbitals
C. two bonding molecular orbitals and two antibonding molecular orbitals D. two bonding molecular orbitals and one antibonding molecular orbital E. one bonding molecular orbital and one antibonding molecular orbital
30. Molecular Orbital theory correctly predicts diamagnetism of fluorine gas, F2. This is because
__________.
A. the bond order of F2 can be shown to be equal to 1
B. there are more electrons in the bonding orbitals than in the antibonding orbitals C. all electrons in the MO electron configuration of F2 are unpaired.
D. the energy of the π2p MOs is higher than that of the σ2p MO E. the F-F bond enthalpy is very low.
SECTION II: Free-response: PLEASE SHOW ALL WORK NEATLY ON A SEPARATE PIECE OF PAPER! Writings on this paper will not be graded!
Part A: Calculator permitted (25 minutes)
Directions:Clearly show the method used and the steps involved in arriving at your answers.
It is to your advantage to do this, since you may obtain partial credit if you do, and you will receive little or no credit if you do not. Attention should be paid to significant figures!
Be sure to write you answers on the separate sheet of lined paper.
1.) The compound chloral hydrate, known in detective stories as knockout drops, is composed of 14.52% C, 1.83% H, 64.30% Cl, and 19.35% O by mass and has a molar mass of 165.4 g/mol.
A. What is the empirical formula of the compound? B. What is the molecular formula of this substance?
C. Draw the Lewis structure of the molecule assuming that the Cl atoms bond to a single C atom, and that there is a C-C bond and two C-O bonds in the compound.
2.) Use the bond enthalpies, electron affinities, and the ionization energy of hydrogen (1312 kJ/mol) to estimate the energy change for the following gas-phase ionization reactions.
Bond Enthalpies: Electron Affinities: H-F 567 kJ/mol F -328 kJ/mol H-Cl 431 kJ/mol Cl -349 kJ/mol H-Br 366 kJ/mol Br -325 kJ/mol
A. HF(g) H+(g) + F-(g) B. HCl(g) H+(g)+ Cl-(g) C. HBr(g) H+(g) + Br-(g)
3.) Boron trifluoride is the chemical compound with the formula BF3. This pungent colorless toxic
gas forms white fumes in moist air.
A. Predict the geometry of BF3.
BF3 is manufactured by the reaction of boron oxides with hydrogen fluoride:
B. Calculate the ΔHrxn of the reaction shown above.
Bond enthalpies (kJ/mol):
B=O 636 B-F 613
B-O 536 H-F 567
H-O 463
C. Assign each of the atoms in the above equation an oxidation number.
Part B: Calculators NOT permitted. (30 min)
______________________________________________________________________________________
Directions: Be sure to write you answers on the separate sheet of lined paper.
1. NF3 and PF5 are stable molecules. Write the electron-dot formulas for these molecules. On the
basis of structural and bonding considerations, account for the fact that NF3 and PF5 are stable
molecules but NF5 does not exist.
2. Draw Lewis structures for CO2, H2S, SO3 and SO32- and predict the shape of each species.
3. The boiling points of the following compounds increase in the order in which they are listed below:
CH4 < H2S < NH3
Discuss the theoretical considerations involved and use them to account for this order.
4. The values of the first three ionization energies (I1, I2, I3) for magnesium and argon are as follows:
I1 I2 I3
(kJ/mol)
Mg 735 1443 7730
Ar 1525 2665 3945
(a) Give the electronic configurations of Mg and Ar.
(b) In terms of these configurations, explain why the values of the first and second ionization energies of Mg are significantly lower than the values for Ar, whereas the third ionization energy of Mg is much larger than the third ionization energy of Ar.
(c) If a sample of Ar in one container and a sample of Mg in another container are each heated and chlorine is passed into each container, what compounds, if any, will be formed? Explain in terms of the electronic configurations given in part (a).
(d) Element Q has the following first three ionization energies:
What is the formula for the most likely compound of
element Q with chlorine? Explain the choice of formula on the basis of the ionization energies.
STOP!
I1 I2 I3
(kJ/mol)
Answer Section
MULTIPLE CHOICE
1. C 2. A 3. E 4. D 5. A 6. B 7. D 8. B 9. B 10. C 11. C 12. D 13. D 14. C 15. D 16. D 17. D 18. D 19. D 20. D 21. D 22. A 23. B 24. C 25. E 26. A 27. A 28. D 29. E 30. A
Free Response Questions Part A:
1.
a) Assume 100g
14.52g C x 1 mole/ 12.01g =1.209 mole C; 1.209/1.209=1 1.83g H x 1mole/ 1.008g= 1.816 mole H; 1.816/1.209= 1.5 64.30g Cl x 1 mole/ 35.45g = 1.814 mole Cl; 1.814/1.209= 1.5 19.35g O x 1 mole/16.00g= 1.209 mole O; 1.209/1.209= 1
b) the empirical formula weight is 2(12.0g)+ 3(1.0g) + 3(35.5g) + 2(16.0g)= 165.5g/mole c) 44 e- total
2C + 3H+ 3Cl+ 2 O
2(4e-)+ 3(1e-) + 3(7e-) + 2(6e-)= 44 e
2.
a) ΔH= 1312 kJ/mol + 567 kJ/mol – 328 kJ/mol = 1551 kJ/mol b) ΔH= 1312 kJ/mol + 431 kJ/mol – 349 kJ/mol = 1394 kJ/mol c) ΔH= 1312 kJ/mol + 366 kJ/mol – 325 kJ/mol = 1353 kJ/mol
3.
a) trigonal planar
b) ΔHrxn = [2(B=O) + 2(B-O) + 6(H-F)] –[6(B-F) + 6(H-O)]
ΔHrxn = [2(636) + 2(536) + 6(567)] –[6(613) + 6(463)]= -710 kJ/mol
c)
B2O3 + 6HF 2BF3 + 3H2O
B=+3 O=-2 H=+1 F=-1 B=+3 F=-3 H=+1 O=-2 Part B 1. . . N . . F . . . . . . . . F
F . . . . . . . . P . . . . . . F . . F . . . . F . . . . . . . . F . .
. . . . . . F . .
Describe the sp3 bonding for NF
3 and the sp3d for PF5.
Nonexistence of NF5 because of no d orbital to involve in bonding.
2. : : : : : : O C O Linear or straight molecule : :
H S Bent or angular molecule H
: : : :
O S O Trigonal pyramidal or dis- torted tetrahedral ion : : O : :
:
: : : : : : : :
2-3. CH4 - weak London dispersion (van der Waals) forces
H2S - London forces + dipole-dipole interactions
NH3 - London + dipole + hydrogen bonding
4. (a) Mg: 1s2 2s22p6 3s1
Ar: 1s2 2s22p6 3s23p6
(b) Valence electrons for Mg and Ar are in the same principal energy level, but Ar atom is smaller and has a greater nuclear charge. Thus, ionization energies for Mg are less than those
for Ar. Removal of third electron from Mg atom is from n = 2 level and electrons in this level
experience strong nuclear attraction.
(c) Only MgCl2 forms. Mg atoms readily lose 2 valence electrons each. Ionization energy for
third electron very high. Electron affinity for Ar is low, and ionization energies for Ar atoms are high.
